Problem 39 - Applied Calculus

Question

Decide if the improper integral \(\int_{0}^{\infty} e^{-2 t} d t\) converges, and if so, to what value, by the following method. (a) Evaluate \(\int_{0}^{b} e^{-2 t} d t\) for \(b=3,5,7,10\). What do you observe? Make a guess about the convergence of the improper integral. (b) Find \(\int_{0}^{b} e^{-2 t} d t\) using the Fundamental Theorem. Your answer will contain \(b\). (c) Take a limit as \(b \rightarrow \infty\). Does your answer confirm your guess?

Step-by-Step Solution

Answer

The improper integral converges to 1/2.

Step 1: Evaluate the Definite Integrals

Calculate \(\int_{0}^{b} e^{-2t} \, dt\) for given values of \(b\): 1. For \(b = 3\), the integral is \([-(1/2)e^{-2t}]_{0}^{3}=(1/2)(1-e^{-6})\).2. For \(b = 5\), the integral is \([-(1/2)e^{-2t}]_{0}^{5}=(1/2)(1-e^{-10})\).3. For \(b = 7\), the integral is \([-(1/2)e^{-2t}]_{0}^{7}=(1/2)(1-e^{-14})\).4. For \(b = 10\), the integral is \([-(1/2)e^{-2t}]_{0}^{10}=(1/2)(1-e^{-20})\).In each case, as \(b\) increases, the value of \(e^{-2b}\) decreases rapidly towards zero, suggesting convergence.

Step 2: Determine the Antiderivative

Apply the Fundamental Theorem of Calculus to find the antiderivative of the function.The antiderivative of \(e^{-2t}\) is \(-(1/2)e^{-2t}\). Thus, the definite integral is:\[ \int_{0}^{b} e^{-2t} \, dt = \left[ -(1/2)e^{-2t} \right]_{0}^{b} = (1/2)(1 - e^{-2b}). \]

Step 3: Evaluate the Limit

Evaluate the limit of the expression \((1/2)(1 - e^{-2b})\) as \(b\) approaches infinity.\[ \lim_{b \to \infty} (1/2)(1 - e^{-2b}) = (1/2)\cdot(1 - 0) = 1/2. \]

Step 4: Confirm Observations and Conclusion

The limit confirms that as \(b\) tends to infinity, the integral converges to \(1/2\). This result aligns with the observations made in Step 1.

Key Concepts

Convergence of Integrals

Improper integrals can extend to infinity or have an infinite discontinuity. To determine if such an integral converges, we need to examine its behavior as it approaches infinity. The convergence of integrals involves checking whether the calculated area remains finite as one or both of the integration bounds move towards infinity.
In this exercise, we calculate the integral \( \int_{0}^{b} e^{-2t} \, dt \) for various values of \( b \), such as 3, 5, 7, and 10. A pattern emerges: as \( b \) increases, the portion of the integral beyond a certain point becomes minuscule because \( e^{-2b} \) shrinks rapidly towards zero. This suggests that the integral is converging to a specific value rather than growing indefinitely.
Convergence is essential as it tells us that the integral has a finite value despite extending over an infinite interval.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration. It enables us to compute a definite integral, which is the accumulated sum of areas, using an antiderivative.
For the function \( e^{-2t} \), we first find its antiderivative, which can be found as \( -(1/2)e^{-2t} \). By applying the theorem, we can evaluate the definite integral from 0 to \( b \):

• Begin with the antiderivative: \( -(1/2)e^{-2t} \)
• Apply the limits of integration 0 to \( b \): \( \left[ -(1/2)e^{-2t} \right]_{0}^{b} \)
• Simplify to obtain \( (1/2)(1 - e^{-2b}) \)

The result implies that the definite integral can be expressed in terms of \( b \), making it easier to evaluate, even as \( b \) becomes very large.

Limit Evaluation

At this stage, we aim to comprehend the behavior of the function expression \((1/2)(1 - e^{-2b})\) as \(b\) approaches infinity. Evaluating limits helps us determine if an expression settles to a particular value, which in terms of integrals, indicates convergence.
By observing the expression:

• As \(b\) increases, \(e^{-2b}\) steadily shrinks toward zero, since any number raised to a larger negative power approaches zero.
• The limit of \((1/2)(1 - e^{-2b})\) is found by taking the limit as \(b\) approaches infinity: \( \lim_{b \to \infty} (1/2)(1 - e^{-2b}) = (1/2)(1 - 0) = 1/2 \).

This clear resolution through limits confirms our hypothesis that the integral converges to \(1/2\), proving that even as the integral stretches infinitely, it remains finite.

Definite Integrals

A definite integral calculates the net area between a function and the x-axis over a specified interval. It becomes more than just calculating areas; it's about understanding how functions behave over intervals.
In this case, we evaluated the definite integral \( \int_{0}^{b} e^{-2t} \, dt \) to identify its bounded nature over the given limits. Calculating definite integrals, such as for values of \(b = 3, 5, 7, 10 \), provides insight into part (a) of the exercise, where we notice the incremental decrease in the exponential term, \( e^{-2b} \).
The realization that these definite integrals approach a consistent total when summed implies that the original improper integral doesn't diverge to infinity but gracefully totals to a finite value, 1/2, as \(b\) approaches infinity.